Deriving the Simple Linear Regression Estimator For the population model described by y = \beta_0 + \beta_1x + \mu obtain the OLS (Ordinary Least Squares) estimator of the intercept, \hat{\beta}_0, and the slope, \hat{\beta}_1, 1. For the Least Squares Method, explain in detail the idea behind the method, and show clearly and precisely all of the steps of the derivations. 2. For the Method of Moments, explain in detail the idea behind the method, and show clearly and precisely all of the steps of the derivations.
The Correct Answer and Explanation is:
Deriving the Simple Linear Regression Estimator
Population Model:
We begin with the simple linear regression population model:y=β0+β1x+μy = \beta_0 + \beta_1 x + \muy=β0+β1x+μ
where:
- yyy is the dependent variable
- xxx is the independent variable
- β0\beta_0β0 is the intercept
- β1\beta_1β1 is the slope
- μ\muμ is the error term
We observe nnn data points: (xi,yi)(x_i, y_i)(xi,yi) for i=1,2,…,ni = 1, 2, \dots, ni=1,2,…,n
1. Ordinary Least Squares (OLS) Estimation
Idea Behind the Method:
The Least Squares Method minimizes the sum of squared residuals (SSR). A residual is the difference between the observed value yiy_iyi and the predicted value y^i=β^0+β^1xi\hat{y}_i = \hat{\beta}_0 + \hat{\beta}_1 x_iy^i=β^0+β^1xi.SSR=∑i=1n(yi−β^0−β^1xi)2\text{SSR} = \sum_{i=1}^n (y_i – \hat{\beta}_0 – \hat{\beta}_1 x_i)^2SSR=i=1∑n(yi−β^0−β^1xi)2
Steps of Derivation:
We minimize SSR with respect to β^0\hat{\beta}_0β^0 and β^1\hat{\beta}_1β^1:
Step 1: Take partial derivatives∂SSR∂β^0=−2∑(yi−β^0−β^1xi)\frac{\partial \text{SSR}}{\partial \hat{\beta}_0} = -2 \sum (y_i – \hat{\beta}_0 – \hat{\beta}_1 x_i)∂β^0∂SSR=−2∑(yi−β^0−β^1xi)∂SSR∂β^1=−2∑xi(yi−β^0−β^1xi)\frac{\partial \text{SSR}}{\partial \hat{\beta}_1} = -2 \sum x_i (y_i – \hat{\beta}_0 – \hat{\beta}_1 x_i)∂β^1∂SSR=−2∑xi(yi−β^0−β^1xi)
Step 2: Set derivatives equal to zero (first-order conditions):∑(yi−β^0−β^1xi)=0(1)\sum (y_i – \hat{\beta}_0 – \hat{\beta}_1 x_i) = 0 \tag{1}∑(yi−β^0−β^1xi)=0(1)∑xi(yi−β^0−β^1xi)=0(2)\sum x_i (y_i – \hat{\beta}_0 – \hat{\beta}_1 x_i) = 0 \tag{2}∑xi(yi−β^0−β^1xi)=0(2)
Step 3: Solve the system of equations
From (1):nβ^0+β^1∑xi=∑yi⇒β^0=yˉ−β^1xˉn\hat{\beta}_0 + \hat{\beta}_1 \sum x_i = \sum y_i \Rightarrow \hat{\beta}_0 = \bar{y} – \hat{\beta}_1 \bar{x}nβ^0+β^1∑xi=∑yi⇒β^0=yˉ−β^1xˉ
Substitute into (2) and solve for β^1\hat{\beta}_1β^1:β^1=∑(xi−xˉ)(yi−yˉ)∑(xi−xˉ)2\hat{\beta}_1 = \frac{\sum (x_i – \bar{x})(y_i – \bar{y})}{\sum (x_i – \bar{x})^2}β^1=∑(xi−xˉ)2∑(xi−xˉ)(yi−yˉ)
Then plug back to get β^0=yˉ−β^1xˉ\hat{\beta}_0 = \bar{y} – \hat{\beta}_1 \bar{x}β^0=yˉ−β^1xˉ
2. Method of Moments Estimation
Idea Behind the Method:
The Method of Moments sets the population moments equal to their sample counterparts. For simple linear regression, we focus on matching the population expectations and covariances.
From the model:y=β0+β1x+μ⇒E[y]=β0+β1E[x]y = \beta_0 + \beta_1 x + \mu \Rightarrow E[y] = \beta_0 + \beta_1 E[x]y=β0+β1x+μ⇒E[y]=β0+β1E[x]Cov(x,y)=Cov(x,β0+β1x+μ)=β1Var(x)Cov(x, y) = Cov(x, \beta_0 + \beta_1 x + \mu) = \beta_1 Var(x)Cov(x,y)=Cov(x,β0+β1x+μ)=β1Var(x)
Thus:β1=Cov(x,y)Var(x),β0=E[y]−β1E[x]\beta_1 = \frac{Cov(x, y)}{Var(x)}, \quad \beta_0 = E[y] – \beta_1 E[x]β1=Var(x)Cov(x,y),β0=E[y]−β1E[x]
Replace expectations and variances with sample versions:β^1=∑(xi−xˉ)(yi−yˉ)∑(xi−xˉ)2,β^0=yˉ−β^1xˉ\hat{\beta}_1 = \frac{\sum (x_i – \bar{x})(y_i – \bar{y})}{\sum (x_i – \bar{x})^2}, \quad \hat{\beta}_0 = \bar{y} – \hat{\beta}_1 \bar{x}β^1=∑(xi−xˉ)2∑(xi−xˉ)(yi−yˉ),β^0=yˉ−β^1xˉ
This produces the same result as OLS because both rely on matching theoretical and empirical moments.
